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    Can anyone please help. The question is;

    If the exact value of cosx is 1/√5 find the exact value of cos2x.

    Any help will be greatly appreciated. Thanks.
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    (Original post by Bob123457)
    Can anyone please help. The question is;

    If the exact value of cosx is 1/√5 find the exact value of cos2x.

    Any help will be greatly appreciated. Thanks.
    Do the inverse (cos^-1) on 1/√5. This gives you x.

    Now multiply x by 2 and find the cos of that. Done!
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    (Original post by Mehru1214)
    Do the inverse (cos^-1) on 1/√5. This gives you x.

    Now multiply x by 2 and find the cos of that. Done!

    Thanks👍🏻
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    No problem
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    Great
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    (Original post by Bob123457)
    Thanks👍🏻
    Fairly sure it wants you to do this without a calculator... otherwise it's a pretty pointless question tbh
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    If cosx = 1/5, then arccos 1/5 = 1.107149718
    We need to find cos2x, so we do ^*2 = 2.214297436
    We then do the cosine of that, which is -0.6

    This would be on a calculator paper; you wouldn't be expected to do this in your head
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    RDKGames is right. The whole point of this question is to use trigonometric identities otherwise it's just mindless calculator inputting.

    Do you know the 'double angle formula' for the cosine function? It's something you'll need to remember for the future.

    cos(2x) = (cosx)^2 - (sinx)^2 = 2(cosx)^2 - 1

    So, since you know the value of cos(x), you can input that into the above formula to find that cos(2x) = 2*(1/5) - 1 = -3/5
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    (Original post by Mehru1214)
    Do the inverse (cos^-1) on 1/√5. This gives you x.

    Now multiply x by 2 and find the cos of that. Done!
    Spoiler:
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    Have you done my paper?
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    (Original post by RDKGames)
    Fairly sure it wants you to do this without a calculator... otherwise it's a pretty pointless question tbh
    You have to do pointless questions like these on the new spec.
 
 
 
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